Prove the following identities :

sec A(1 - sin A)(sec A + tan A) = 1

Solving L.H.S. of the equation :

$\Rightarrow \text{sec A(1 - sin A)(sec A + tan A)} \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\dfrac{1}{\text{cos A}} + \dfrac{\text{sin A}}{\text{cos A}}\Big) \\[1em] \Rightarrow \dfrac{1}{\text{cos A}} \times (\text{1 - sin A}) \times \Big(\dfrac{\text{1 + sin A}}{\text{cos A}}\Big) \\[1em] \Rightarrow \dfrac{1 - \text{sin}^2 A}{\text{cos}^2A}$

By formula,

cos² A = 1 - sin² A

$\Rightarrow \dfrac{1 - \text{sin}^2 A}{1 - \text{sin}^2A} \\[1em] \Rightarrow 1.$

Since, L.H.S. = R.H.S.

Hence, proved that sec A(1 - sin A)(sec A + tan A) = 1.